edge = resubEdge(tree)
returns
the classification edge obtained by edge
= resubEdge(tree
)tree
on its
training data.

A classification tree created using 

Classification edge obtained by resubstituting the training data into the calculation of edge. 
The edge is the weighted mean value of
the classification margin. The weights are the class probabilities
in tree
.Prior
.
The classification margin is the difference
between the classification score for the true
class and maximal classification score for the false classes. Margin
is a column vector with the same number of rows as in the matrix X
.
For trees, the score of a classification of a leaf node is the posterior probability of the classification at that node. The posterior probability of the classification at a node is the number of training sequences that lead to that node with the classification, divided by the number of training sequences that lead to that node.
For example, consider classifying a predictor X
as true
when X
< 0.15
or X
> 0.95
, and X
is
false otherwise.
Generate 100 random points and classify them:
rng(0,'twister') % for reproducibility X = rand(100,1); Y = (abs(X  .55) > .4); tree = fitctree(X,Y); view(tree,'Mode','Graph')
Prune the tree:
tree1 = prune(tree,'Level',1); view(tree1,'Mode','Graph')
The pruned tree correctly classifies observations that are less
than 0.15 as true
. It also correctly classifies
observations from .15 to .94 as false
. However,
it incorrectly classifies observations that are greater than .94 as false
.
Therefore, the score for observations that are greater than .15 should
be about .05/.85=.06 for true
, and about .8/.85=.94
for false
.
Compute the prediction scores for the first 10 rows of X
:
[~,score] = predict(tree1,X(1:10)); [score X(1:10,:)]
ans = 0.9059 0.0941 0.8147 0.9059 0.0941 0.9058 0 1.0000 0.1270 0.9059 0.0941 0.9134 0.9059 0.0941 0.6324 0 1.0000 0.0975 0.9059 0.0941 0.2785 0.9059 0.0941 0.5469 0.9059 0.0941 0.9575 0.9059 0.0941 0.9649
Indeed, every value of X
(the rightmost
column) that is less than 0.15 has associated scores (the left and
center columns) of 0
and 1
,
while the other values of X
have associated scores
of 0.91
and 0.09
. The difference
(score 0.09
instead of the expected .06
)
is due to a statistical fluctuation: there are 8
observations
in X
in the range (.95,1)
instead
of the expected 5
observations.
Estimate the quality of a classification tree for the Fisher iris data by resubstitution.
load fisheriris
tree = fitctree(meas,species);
redge = resubEdge(tree)
redge = 0.9384
edge
 fitctree
 resubLoss
 resubMargin
 resubPredict